CHAPTER 7: Linear Momentum -...

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

164

CHAPTER 7: Linear Momentum

Answers to Questions

1. For momentum to be conserved, the system under analysis must be closed not have any forces on it from outside the system. A coasting car has air friction and road friction on it, for example, which are outside forces and thus reduce the momentum of the car. If the ground and the air were considered part of the system, and their velocities analyzed, then the momentum of the entire system would be conserved, but not necessarily the momentum of any single component, like the car.

2. Consider this problem as a very light object hitting and sticking to a very heavy object. The large object small object combination (Earth + jumper) would have some momentum after the collision, but due to the very large mass of the Earth, the velocity of the combination is so small that it is not measurable. Thus the jumper lands on the Earth, and nothing more happens.

3. When you release an inflated but untied balloon at rest, the gas inside the balloon (at high pressure) rushes out the open end of the balloon. That escaping gas and the balloon form a closed system, and so the momentum of the system is conserved. The balloon and remaining gas acquires a momentum equal and opposite to the momentum of the escaping gas, and so move in the opposite direction to the escaping gas.

4. If the rich man would have faced away from the shore and thrown the bag of coins directly away from the shore, he would have acquired a velocity towards the shore by conservation of momentum. Since the ice is frictionless, he would slide all the way to the shore.

5. When a rocket expels gas in a given direction, it puts a force on that gas. The momentum of the gas-rocket system stays constant, and so if the gas is pushed to the left, the rocket will be pushed to the right due to Newton s 3rd law. So the rocket must carry some kind of material to be ejected (usually exhaust from some kind of engine) to change direction.

6. The air bag greatly increases the amount of time over which the stopping force acts on the driver. If a hard object like a steering wheel or windshield is what stops the driver, then a large force is exerted over a very short time. If a soft object like an air bag stops the driver, then a much smaller force is exerted over a much longer time. For instance, if the air bag is able to increase the time of stopping by a factor of 10, then the average force on the person will be decreased by a factor of 10. This greatly reduces the possibility of serious injury or death.

7. Crumple zones are similar to air bags in that they increase the time of interaction during a collision, and therefore lower the average force required for the change in momentum that the car undergoes in the collision.

8. From Eq. 7-7 for a 1-D elastic collision, A B B Av v v v . Let A represent the bat, and let B

represent the ball. The positive direction will be the (assumed horizontal) direction that the bat is moving when the ball is hit. We assume the batter can swing the bat with equal strength in either case, so that Av is the same in both pitching situations. Because the bat is so much heavier than the

ball, we assume that A Av v

the speed of the bat doesn t change significantly during the collision.

Then the velocity of the baseball after being hit is 2B A A B A Bv v v v v v . If 0Bv , the ball

tossed up into the air by the batter, then 2B Av v

the ball moves away with twice the speed of the

Giancoli Physics: Principles with Applications, 6th Edition


165

bat. But if 0Bv , the pitched ball situation, we see that the magnitude of 2B Av v , and so the ball

moves away with greater speed. If, for example, the pitching speed of the ball was about twice the speed at which the batter could swing the bat, then we would have 4B Av v . Thus the ball has

greater speed after being struck, and thus it is easier to hit a home run. This is similar to the gravitational slingshot effect discussed in problem 85.

9. The impulse is the product of the force and the time duration that the force is applied. So the impulse from a small force applied over a long time can be larger than the impulse applied by a large force over a small time.

10. The momentum of an object can be expressed in terms of its kinetic energy, as follows. 2 2 2 21

22 2p mv m v m mv m mv mKE .

Thus if two objects have the same kinetic energy, then the one with more mass has the greater momentum.

11. Consider two objects, each with the same magnitude of momentum, moving in opposite directions. They have a total momentum of 0. If they collide and have a totally inelastic collision, in which they stick together, then their final common speed must be 0 so that momentum is conserved. But since they are not moving after the collision, they have no kinetic energy, and so all of their kinetic energy has been lost.

12. The turbine blades should be designed so that the water rebounds. If the water rebounds, that means that a larger momentum change for the water has occurred than if it just came to a stop. And if there is a larger momentum change for the water, there will also be a larger momentum change for the blades, making them spin faster.

13. (a) The downward component of the momentum is unchanged. The horizontal component of momentum changes from rightward to leftward. Thus the change in momentum is to the left in the picture.

(b) Since the force on the wall is opposite that on the ball, the force on the wall is to the right.

14. (a) The momentum of the ball is not conserved during any part of the process, because there is an external force acting on the ball at all times the force of gravity. And there is an upward force on the ball during the collision. So considering the ball as the system, there are always external forces on it, and so its momentum is not conserved.

(b) With this definition of the system, all of the forces are internal, and so the momentum of the Earth-ball system is conserved during the entire process.

(c) The answer here is the same as for part (b).

15. In order to maintain balance, your CM must be located directly above your feet. If you have a heavy load in your arms, your CM will be out in front of your body and not above your feet. So you lean backwards to get your CM directly above your feet. Otherwise, you would fall over forwards.

16. The 1-m length of pipe is uniform it has the same density throughout, and so its CM is at its geometric center, which is its midpoint. The arm and leg are not uniform they are more dense where there is muscle, primarily in the parts that are closest to the body. Thus the CM of the arm or leg is closer the body than the geometric center. The CM is located closer to the more massive part of the arm or leg.

Chapter 7 Linear Momentum


166

17. When you are lying flat on the floor, your CM is inside of the volume of your body. When you sit up on the floor with your legs extended, your CM is outside of the volume of your body.

18. The engine does not directly accelerate the car. The engine puts a force on the driving wheels, making them rotate. The wheels then push backwards on the roadway as they spin. The Newton s 3rd law reaction to this force is the forward-pushing of the roadway on the wheels, which accelerates the car. So it is the (external) road surface that accelerates the car.

19. The motion of the center of mass of the rocket will follow the original parabolic path, both before and after explosion. Each individual piece of the rocket will follow a separate path after the explosion, but since the explosion was internal to the system (consisting of the rocket), the center of mass of all the exploded pieces will follow the original path.

Solutions to Problems

1. 0.028 kg 8.4 m s 0.24 kg m sp mv

2. From Newton s second law, tp F . For a constant mass object, mp v . Equate the two expressions for p .

tt m

m

FF v v .

If the skier moves to the right, then the speed will decrease, because the friction force is to the left. 25 N 20 s

7.7 m s65 kg

F tv

m

The skier loses 7.7 m s of speed.

3. Choose the direction from the batter to the pitcher to be the positive direction. Calculate the average force from the change in momentum of the ball.

3

3

52.0 m s 39.0 m s0.145 kg 4.40 10 N, towards the pitcher

3.00 10 s

p F t m v

vF m

t

4. The throwing of the package is a momentum-conserving action, if the water resistance is ignored. Let A represent the boat and child together, and let B represent the package. Choose the direction that the package is thrown as the positive direction. Apply conservation of momentum, with the initial velocity of both objects being 0.

initial final

6.40 kg 10.0 m s0.901m s

26.0 kg 45.0 kg

A B A A B B

B BA

A

p p m m v m v m v

m vv

m

The boat and child move in the opposite direction as the thrown package.

CM CM



167

5. The force on the gas can be found from its change in momentum.

4 74.0 10 m s 1500 kg s 6.0 10 N downwardp v m m

F vt t t

The force on the rocket is the Newton s 3rd law pair (equal and opposite) to the force on the gas, and

so is 76.0 10 N upward .

6. The tackle will be analyzed as a one-dimensional momentum conserving situation. Let A represent the halfback, and B represent the tackling cornerback.

initial final

95 kg 4.1m s 85 kg 5.5 m s4.8 m s

95 kg 85 kg

A A B B A B

A A B B

A B

p p m v m v m m v

m v m vv

m m

7. Consider the horizontal motion of the objects. The momentum in the horizontal direction will be conserved. Let A represent the car, and B represent the load. The positive direction is the direction of the original motion of the car.

initial final

12, 600 kg 18.0 m s 012.6 m s

12, 600 kg 5350 kg

A A B B A B

A A B B

A B

p p m v m v m m v

m v m vv

m m

8. Consider the motion in one dimension, with the positive direction being the direction of motion of the first car. Let A represent the first car, and B represent the second car. Momentum will be conserved in the collision. Note that 0Bv .

initial final

4 9300 kg 15.0 m s 6.0 m s1.4 10 kg

6.0 m s

A A B B A B

A AB

p p m v m v m m v

m v vm

v

9. The force stopping the wind is exerted by the person, so the force on the person would be equal in magnitude and opposite in direction to the force stopping the wind. Calculate the force from Eq. 7-2, in magnitude only.

windwind2

wind wind wind windon on windperson wind

2

40 kg s 1m s1.50 m 0.50 m 30 kg s 100 km h 27.8 m s

m 3.6 km h

30 kg s 27.8 m s

833 N 8 10 N

mv

t

p m v mF F v

t t t

The typical maximum frictional force is 2

fr 1.0 70 kg 9.8 m s 690 NsF mg , and so we

see that on frperson

F F

the wind is literally strong enough to blow a person off his feet.

10. Momentum will be conserved in the horizontal direction. Let A represent the car, and B represent the snow. For the horizontal motion, 0Bv and B Av v . Momentum conservation gives

the following.



168

initial final

3800 kg 8.60 m s7.94 m s 7.9 m s

3.50 kg3800 kg 90.0 min

min

A A A B A

A AA

A B

p p m v m m v

m vv

m m

11. Consider the motion in one dimension, with the positive direction being the direction of motion of

the original nucleus. Let A represent the alpha particle, with a mass of 4 u, and B represent the new nucleus, with a mass of 218 u. Momentum conservation gives the following.

initial final

3 222 u 420 m s 218 u 350 m s4.2 10 m s

4.0 u

A B A A B B

A B B BA

A

p p m m v m v m v

m m v m vv

m

Note that the masses do not have to be converted to kg, since all masses are in the same units, and a ratio of masses is what is significant.

12. Consider the motion in one dimension with the positive direction being the direction of motion of the bullet. Let A represent the bullet, and B represent the block. Since there is no net force outside of the block-bullet system (like frictions with the table), the momentum of the block and bullet combination is conserved. Note that 0Bv .

initial final

0.023 kg 230 m s 170 m s0.69 m s

2.0 kg

A A B B A A B B

A A AB

B

p p m v m v m v m v

m v vv

m

13. (a) Consider the motion in one dimension with the positive direction being the direction of motion before the separation. Let A represent the upper stage (that moves away faster) and B represent the lower stage. It is given that A Bm m , A Bv v v , and B A relv v v . Momentum

conservation gives the following.

initial final

12

3

3 3 3

975 kg 5800 m s 975 kg 2200 m s

975 kg

6.9 10 m s , away from Earth

6.9 10 m s 2.20 10 m s 4.7 10 m s , away fro

A B A A B B A A B A rel

A B B relA

A B

B A rel

p p m m v m v m v m v m v v

m m v m vv

m m

v v v m Earth

(b) The change in KE had to be supplied by the explosion.

2 2 21 1 12 2 2

2 2 21 12 2

8

487.5kg 6900 m s 4700 m s 975 kg 5800 m s

5.9 10 J

f i A A B B A BKE KE KE m v m v m m v



169

14. To alter the course by 35.0o, a velocity perpendicular to the original velocity must be added. Call the direction of the added velocity, addv , the positive

direction. From the diagram, we see that add orig tanv v . The momentum in

the perpendicular direction will be conserved, considering that the gases are given perpendicular momentum in the opposite direction of addv . The gas is

expelled in the opposite direction to addv , and so a negative value is used for

gasv .

gas gas rocket gas addbefore after

o2rocket add

gas oadd gas

0

3180 kg 115 m s tan 35.01.40 10 kg

115 m s tan 35 1750 m s

p p m v m m v

m vm

v v

15. (a) The impulse is the change in momentum. The direction of travel of the struck ball is the positive direction.

24.5 10 kg 45m s 0 2.0 kg m sp m v

(b) The average force is the impulse divided by the interaction time.

2

3

2.0 kg m s5.8 10 N

3.5 10 s

pF

t

16. (a) The impulse given to the nail is the opposite of the impulse given to the hammer. This is the change in momentum. Call the direction of the initial velocity of the hammer the positive direction.

2

nail hammer 12 kg 8.5 m s 0 1.0 10 kg m si fp p mv mv

(b) The average force is the impulse divided by the time of contact. 2

4

avg 3

1.0 10 kg m s1.3 10 N

8.0 10 s

pF

t

17. The impulse given the ball is the change in the ball s momentum. From the symmetry of the problem, the vertical momentum of the ball does not change, and so there is no vertical impulse. Call the direction AWAY from the wall the positive direction for momentum perpendicular to the wall.

o o o

final initial

2 o

sin 45 sin 45 2 sin 45

2 6.0 10 km 25 m s sin 45 2.1kg m s , to the left

p mv mv m v v mv

18. (a) The average force on the car is the impulse (change in momentum) divided by the time of interaction. The positive direction is the direction of the car s initial velocity.

5 5

1m s0 50 km h

3.6 km h1500 kg 1.389 10 N 1.4 10 N

0.15 s

p m vF

t t

origv

addv

finalv



170

(b) The deceleration is found from Newton s 2nd law. 5

21.389 10 N

93m s

1500 kg

FF ma a

m

19. Call east the positive direction.

(a) 2

original originalfullback fullback

95 kg 4.0 m s 3.8 10 kg m sp mv

(b) The impulse on the fullback is the change in the fullback s momentum.

2

fullback final finalfullback fullback

95 kg 0 4.0 m s 3.8 10 kg m sp m v v

(c) The impulse on the tackler is the opposite of the impulse on the fullback, so 23.8 10 kg m s

(d) The average force on the tackler is the impulse on the tackler divided by the time of interaction. 2

23.8 10 kg m s5.1 10 N

0.75 s

pF

t

20. (a) The impulse given the ball is the area under the F vs. t graph. Approximate the area as a triangle of height 250 N, and width 0.01 sec.

12

250 N 0.01 s 1.25N sp

(b) The velocity can be found from the change in momentum. Call the positive direction the direction of the ball s travel after being served.

i -2

1.25N s

0 21m s6.0 10 kg

f f i

pp m v m v v v v

m

21. Find the velocity upon reaching the ground from energy conservation. Assume that all of the initial potential energy at the maximum height maxh is converted into kinetic energy. Take down to be the

positive direction, so the velocity at the ground is positive. 21

max ground ground max2 2mgh mv v gh

When contacting the ground, the impulse on the person causes a change in momentum. That relationship is used to find the time of the stopping interaction. The force of the ground acting on the person is negative since it acts in the upward direction.

ground

ground0 mv

F t m v tF

We assume that the stopping force is so large that we call it the total force on the person we ignore gravity for the stopping motion. The average acceleration of the person during stopping a F m

is used with Eq. 2-11b to find the displacement during stopping, stoph . 2

ground ground21 10 0 stop ground2 2

2 2 2

ground ground ground stopmax1 1stop max2 2

mv mvFy y v t at h v

F m F

mv mv mv Fhgh mh h

F F F F mg



171

We assume that the person lands with both feet striking the ground simultaneously, so the stopping force is divided between both legs. Thus the critical average stopping force is twice the breaking strength of a single leg.

6 2 4 2

stop break stop

max 2

2 170 10 N m 2.5 10 m 0.60 m269 m

75 kg 9.8 m s

Fh F hh

mg mg

22. Let A represent the 0.440-kg ball, and B represent the 0.220-kg ball. We have A 3.30 m sv and

B 0v . Use Eq. 7-7 to obtain a relationship between the velocities.

A B A B B A A

v v v v v v v

Substitute this relationship into the momentum conservation equation for the collision.

A A B B A A B B A A A A B A A

A BA A

A B

B A A

0.220 kg3.30 m s 1.10 m s east

0.660 kg

3.30 m s 1.10 m s 4.40 m s east

m v m v m v m v m v m v m v v

m mv v

m m

v v v

23. Let A represent the 0.450-kg puck, and let B represent the 0.900-kg puck. The initial direction of

puck A is the positive direction. We have A 3.00 m sv and B 0v . Use Eq. 7-7 to obtain a

relationship between the velocities.

A B A B B A A

v v v v v v v



A BA A

A B

B A A

0.450 kg3.00 m s 1.00 m s 1.00 m s west

1.350 kg

3.00 m s 1.00 m s 2.00 m s east


m mv v

m m

v v v

24. Let A represent the ball moving at 2.00 m/s, and call that direction the positive direction. Let B represent the ball moving at 3.00 m/s in the opposite direction. So A 2.00 m sv and

B 3.00 m sv . Use Eq. 7-7 to obtain a relationship between the velocities.

A B A B B A

5.00 m sv v v v v v

Substitute this relationship into the momentum conservation equation for the collision, noting that

A Bm m .

A A B B A A B B A B A B

A A A A

B A

1.00 m s 5.00 m s 2 6.00 m s 3.00 m s

5.00 m s 2.00 m s

m v m v m v m v v v v v

v v v v

v v

The two balls have exchanged velocities. This will always be true for 1-D elastic collisions of objects of equal mass.



172

25. Let A represent the 0.060-kg tennis ball, and let B represent the 0.090-kg ball. The initial direction of the balls is the positive direction. We have A 2.50 m sv and B 1.15 m sv . Use Eq. 7-7 to

obtain a relationship between the velocities.

A B A B B A

1.35m sv v v v v v


A A B B A A B B A A B B A A B A

A A B BA

A B

B A

1.35 m s

1.35 m s 0.060 kg 2.50 m s 0.090 kg 1.15 m s 1.35 m s

0.150 kg

0.88 m s

1.35 m s 2.23m s

m v m v m v m v m v m v m v m v

m v m vv

m m

v v

Both balls move in the direction of the tennis ball s initial motion.

26. Let A represent the moving softball, and let B represent the ball initially at rest. The initial direction of the softball is the positive direction. We have A 8.5m sv , B 0v , and A 3.7 m sv .

(a) Use Eq. 7-7 to obtain a relationship between the velocities.

A B A B B A B A

8.5m s 0 3.7 m s 4.8m sv v v v v v v v

(b) Use momentum conservation to solve for the mass of the target ball.

A A B B A A B B

A AB A

B B

8.5m s 3.7 m s0.220 kg 0.56 kg

4.8m s

m v m v m v m v

v vm m

v v

27. Let the original direction of the cars be the positive direction. We have A 4.50 m sv and

B 3.70 m sv

(a) Use Eq. 7-7 to obtain a relationship between the velocities.

A B A B B A B A A

0.80 m sv v v v v v v v v


A A B B A A B B A A B B A A B A

A A B BA

A B

B A

0.80 m s

0.80 m s 450 kg 4.50 m s 550 kg 2.90 m s3.62 m s

1000 kg

0.80 m s 4.42 m s

m v m v m v m v m v m v m v m v

m v m vv

m m

v v

(b) Calculate p p p for each car. 2

A A A A A

2

2

B B B B B

2

450 kg 3.62 m s 4.50 m s 3.96 10 kg m s

4.0 10 kg m s

550 kg 4.42 m s 3.70 m s 3.96 10 kg m s

4.0 10 kg m s

p m v m v

p m v m v

The two changes are equal and opposite because momentum was conserved.



173

28. (a) Momentum will be conserved in one dimension. Call the direction of the first ball the positive direction. Let A represent the first ball, and B represent the second ball. We have

0Bv and 12B Av v . Use Eq. 7-7 to obtain a relationship between the velocities.

1A B A B A 2 Av v v v v v


1 1initial final 2 2

A A B B A A B B A A A A B Ap p m v m v m v m v m v m v m v

B A3 3 0.280 kg 0.840 kgm m

(b) The fraction of the kinetic energy given to the second ball is as follows. 22 11

22

2 212

30.75A AB BB

A A A A A

m vm vKE

KE m v m v

29. Let A represent the cube of mass M, and B represent the cube of mass m. Find the speed of A

immediately before the collision, Av , by using energy conservation.

2 21A A2

2 2 9.8m s 0.30 m 2.42 m sMgh Mv v gh

Use Eq. 7-7 for elastic collisions to obtain a relationship between the velocities in the collision. We have B 0v and 2M m .

A B A B B A A

v v v v v v v



2

AA A A A A

4B A A A3

2 9.8 m s 0.30 m22 2 0.808 m s

3 3 3

3.23m s


ghvmv mv m v v v

v v v v

Each mass is moving horizontally initially after the collision, and so each has a vertical velocity of 0 as they start to fall. Use constant acceleration Eq. 2-11b with down as positive and the table top as the vertical origin to find the time of fall.

2 21 10 0 2 2

0 0 2y y v t at H gt t H g

Each cube then travels a horiztonal distance found by xx v t .

2 23 3

8 83 3

2 20.30 m 0.90 m 0.3464 m 0.35 m

3

4 2 20.30 m 0.90 m 1.386 m 1.4 m

3

m A

M B

gh Hx v t hH

g

gh Hx v t hH

g

30. (a) Use Eq. 7-7, along with 0Bv , to obtain a relationship between the velocities.

A B A B B A A

v v v v v v v


A A B B A A B B A A B A A A A B A B A

A BA A B A A A B A A B A A B A A A

A B

m v m v m v m v m v m v v m v m v m v

m mm v m v m v m v m m v m m v v v

m m

Substitute this result into the result of Eq. 7-7.



174

A B A B A B AB A A A A A A A

A B A B A B A B

2m m m m m m mv v v v v v v v

m m m m m m m m

(b) If A Bm m , then approximate 0Am

2

0A B B A A

A A A A B

A B B A B

m m m m vv v v v v

m m m m m

The result is

; 0A A Bv v v . An example of this is a ball bouncing off of the floor. The

massive floor has no speed after the collision, and the velocity of the ball is reversed (if dissipative forces are not present).

(c) If A Bm m , then approximate 0Bm .

2 2

2A B A A A A AA A A A B A

A B A A B A

m m m m v m vv v v v v v

m m m m m m

The result is A

; 2A A Bv v v v . An example of this would be a golf club hitting a golf ball.

The speed of the club immediately after the collision is essentially the same as its speed before the collision, and the golf ball takes off with twice the speed of the club.

(d) If A Bm m , then set A Bm m m .

2 20

2A A

A A B A

m m mv mvv v v v

m m m m m

The result is A0 ; A Bv v v . An example of this is one billiard ball making a head-on

collision with another. The first ball stops, and the second ball takes off with the same speed that the first one had.

31. From the analysis in Example 7-10, the initial projectile speed is given by 2m M

v ghm

.

Compare the two speeds with the same masses.

222 2

2 1

1 111

25.2

2 22.62

m Mgh hv hm v v

m Mv hhghm

32. From the analysis in the Example 7-10, we know that

22

2

2

0.028 kg 230 m s1 1

2 0.028 kg 3.6kg2 9.8m s

0.1607 m 0.16 m

m Mv gh

m

mvh

g m M

From the diagram we see that 22 2

2 2 22 2.8 m 2.8 m 0.1607 m 0.94 m

L L h x

x L L h

L

x

L - h

h



175

33. (a) In example 7-10, 212iKE mv and 21

2fKE m M v . The speeds are related by

mv v

m M.

2

22 21 1

2 2

2 212

2 22

21

f i

i i

mm M v mv

KE KE m M v mvKE m M

KE KE mv mv

m vmv

m Mm Mmv m M m M

(b) For the given values, 380 g

0.96394 g

M

m M. Thus 96% of the energy is lost.

34. Use conservation of momentum in one dimension, since the particles will separate and travel in opposite directions. Call the direction of the heavier particle s motion the positive direction. Let A represent the heavier particle, and B represent the lighter particle. We have A B1.5m m , and

A B 0v v .

B B 2initial final A A B B A B3

A

0 m v

p p m v m v v vm

The negative sign indicates direction.

Since there was no mechanical energy before the explosion, the kinetic energy of the particles after the explosion must equal the energy added.

22 2 2 25 51 1 1 2 1 1added 2 2 2 3 2 3 2 3

3 3added added5 5

1.5

7500 J 4500 J 7500 J 4500 J 3000 J

A B A A B B B B B B B B B

B A B

E KE KE m v m v m v m v m v KE

KE E KE E KE

Thus 3 33.0 10 J 4.5 10 JA BKE KE

35 Use conservation of momentum in one dimension. Call the direction of the sports car s velocity the positive x direction. Let A represent the sports car, and B represent the SUV. We have 0Bv and

A Bv v . Solve for Av .

A Binitial final A A A B A A A

A

0 m m

p p m v m m v v vm

The kinetic energy that the cars have immediately after the collision is lost due to negative work done by friction. The work done by friction can also be calculated using the definition of work. We assume the cars are on a level surface, so that the normal force is equal to the weight. The distance

the cars slide forward is x . Equate the two expressions for the work done by friction, solve for Av ,

and use that to find Av . 21

afterfr final initial A B A2collision

o

fr fr A B

21A B A A B A2

0

cos180

2

k

k k

W KE KE m m v

W F x m m g x

m m v m m g x v g x



176

2A B A BA A

A A

920 kg 2300 kg2 2 0.80 9.8 m s 2.8 m

920 kg

23.191m s 23m s

k

m m m mv v g x

m m

36. Consider conservation of energy during the rising and falling of the ball, between contacts with the

floor. The gravitational potential energy at the top of a path will be equal to the kinetic energy at the

start and the end of each rising-falling cycle. Thus 212

mgh mv for any particular bounce cycle.

Thus for an interaction with the floor, the ratio of the energies before and after the bounce is

after

before

1.20 m0.80

1.50 m

KE mgh

KE mgh. We assume that each bounce will further reduce the energy to

80% of its pre-bounce amount. The number of bounces to lose 90% of the energy can be expressed as follows.

log 0.10.8 0.1 10.3

log 0.8

nn

Thus after 11 bounces, more than 90% of the energy is lost.

As an alternate method, after each bounce, 80% of the available energy is left. So after 1 bounce, 80% of the original energy is left. After the second bounce, only 80% of 80%, or 64% of the available energy is left. After the third bounce, 51 %. After the fourth bounce, 41%. After the fifth bounce, 33 %. After the sixth bounce, 26%. After the seventh bounce, 21%. After the eight bounce, 17%. After the ninth bounce, 13%. After the tenth bounce, 11%. After the eleventh bounce, 9% is left. So again, it takes 11 bounces.

37. (a) For a perfectly elastic collision, Eq. 7-7 says A B A Bv v v v . Substitute that into the

coefficient of restitution definition.

1A BA B

B A B A

v vv ve

v v v v.

For a completely inelastic collision, A Bv v . Substitute that into the coefficient of restitution

definition.

0A B

B A

v ve

v v.

(b) Let A represent the falling object, and B represent the heavy steel plate. The speeds of the steel plate are 0Bv and 0Bv . Thus A Ae v v . Consider energy conservation during the

falling or rising path. The potential energy of body A at height h is transformed into kinetic energy just before it collides with the plate. Choose down to be the positive direction.

21A A2

2mgh mv v gh

The kinetic energy of body A immediately after the collision is transformed into potential energy as it rises. Also, since it is moving upwards, it has a negative velocity.

21A A2

2mgh mv v gh

Substitute the expressions for the velocities into the definition of the coefficient of restitution.

2

2A A

ghe v v e h h

gh



177

38. Let A represent the more massive piece, and B the less massive piece. Thus A B3m m . In the

explosion, momentum is conserved. We have A B 0v v . 1

initial final A A B B B A B B A B3 0 3 p p m v m v m v m v v v

For each block, the kinetic energy gained during the explosion is lost to negative work done by friction on the block.

21fr 2f iW KE KE mv

But work is also calculated in terms of the force doing the work and the distance traveled. o

fr fr Ncos180 k kW F x F x mg x

Equate the two work expressions, solve for the distance traveled, and find the ratio of distances. 2

A222 1

2 B3A A12 2 2 2

B B BB

1

9k

k

k

k

v

x vg vvmv mg x x

vg x v v

g

And so heavy light

1 9x x

39. In each case, use momentum conservation. Let A represent the 15.0-kg object, and let B represent the 10.0-kg object. We have A 5.5 m sv and B 4.0 m sv .

(a) In this case, A Bv v .

A A B B A B A

A A B BB A

A B

15.0 kg 5.5 m s 10.0 kg 4.0 m s1.7 m s

25.0 kg

m v m v m m v

m v m vv v

m m

(b) In this case, use Eq. 7-7 to find a relationship between the velocities.

A B A B B A B A

A A B B A A B B A A B A B A

A B A B BA

A B

B A B A

2 5.0 kg 5.5m s 2 10.0 kg 4.0 m s2.1m s

25.0 kg

5.5m s 4.0 m s 2.1m s 7.4 m s

v v v v v v v v

m v m v m v m v m v m v v v

m m v m vv

m m

v v v v

(c) In this case, A 0v .

A A B B B B

A A B BB

B

15.0 kg 5.5 m s 10.0 kg 4.0 m s4.3m s

10.0 kg

m v m v m v

m v m vv

m

To check for reasonableness , first note the final directions of motion. A has stopped, and B has gone in the opposite direction. This is reasonable. Secondly, calculate the change in kinetic energy.

2 2 21 1 1B B A A B B2 2 2

2 21 1 12 2 2

10.0 kg 15.0 kg 5.5m s 10.0 kg 4.0 m s 220 J

KE m v m v m v

Since the system has lost kinetic energy and the directions are possible, this interaction is reasonable .



178

(d) In this case, B 0v .

A A B B A A

A A B BA

A

15.0 kg 5.5 m s 10.0 kg 4.0 m s2.8 m s

15.0 kg

m v m v m v

m v m vv

m

This answer is not reasonable because it has A moving in its original direction while B has stopped. Thus A has somehow passed through B. If B has stopped, A should have rebounded in the negative direction.

(e) In this case, A 4.0 m sv .

A A B B A A B B m v m v m v m v

B

15.0 kg 5.5m s 4.0 m s 10.0 kg 4.0 m s10.3m s

10.0 kgv

The directions are reasonable, in that each object rebounds. However, the speed of both objects is larger than its speed in the perfectly elastic case (b). Thus the system has gained kinetic energy, and unless there is some other source adding energy, this is not reasonable.

40. Use this diagram for the momenta after the decay. Since there was no momentum before the decay, the three momenta shown must add to 0 in both the x and y directions.

nucleus neutrino nucleus electron

2 2 2 2

nucleus nucleus nucleus neutrino electron

x y

x y

p p p p

p p p p p

2 223 23 22

23nucleus1 1 1 oelectron

23nucleus neutrino

5.40 10 kg m s 9.30 10 kg m s 1.08 10 kg m s

9.30 10 kg m stan tan tan 59.9

5.40 10 kg m sy

x

p p

p p

The second nucleus momentum is 150o from the momentum of the electron.

41. Consider the diagram for the momenta of the eagles. Momentum will be conserved in both the x and y directions.

A AA B A A

A B

B BA B B B

A B

x x x

y y y

m vp m m v m v v

m m

m vp m m v m v v

m m

2 2 2 2

A A B B2 2 A A B B

A B A B A B

2 2 2 24.3 kg 7.8m s 5.6 kg 10.2 m s

6.7 m s4.3 kg 5.6 kg

x y

m v m vm v m vv v v

m m m m m m

electronp

nucleusp

neutrinop

B B Bmp v

A A Amp v

A Bm mp v



179

B B

1 1 1 1 oA B B B

A A A A

A B

5.6 kg 10.2 m stan tan tan tan 60 rel. to eagle A

4.3 kg 7.8m sy

x

m v

v m m m v

m vv m vm m

42. (a) A A A A A B B B: cos cosxp m v m v m v

A A A B B B: 0 sin sinyp m v m v

(b) Solve the x equation for Bcos and the y equation for Bsin , and then

find the angle from the tangent function.

A A A

B B B A AB

A A A AB A A A

B B

sin

sin sintan

coscos cos

m v

m v v

m v v v v

m v

o1 1 o

o

1.10 m s sin 30.0sintan tan 33.0

cos 1.80 m s 1.10 m s cos 30.0A A

B

A A A

v

v v

With the value of the angle, solve the y equation for the velocity. o

A A AB o

B B

0.400 kg 1.10 m s sin 30.0sin0.808 m s

sin 0.500 kg sin 33.0

m vv

m

43. Call the final direction of the joined objects the positive x axis. A diagram of the collision is shown. Momentum will be conserved in both the x and y directions. Note that A Bv v v and 3v v .

1 2 1 2 1 2

21 2 1 2 3

1 o2 11 2 1 1 23 3

o

1 2

: sin sin 0 sin sin

: cos cos 2 3 cos cos

cos cos 2 cos cos 70.5

141

y

x

p mv mv

p mv mv m v

44. Write momentum conservation in the x and y directions, and KE conservation. Note that both masses are the same. We allow Av to have both x and y components.

B A B A

A A B A A B

2 2 2 2 2 2 2 21 1 1 1A B A B A B A B2 2 2 2

:

:

:

x x x

y y y

p mv mv v v

p mv mv mv v v v

KE mv mv mv mv v v v v

Substitute the results from the momentum equations into the KE equation. 2 2 2 2 2 2 2 2 2 2

A B A A B A A B B A A B

2 2 2 2 2 2

A A B B A B A B A B

2

2 2 0 0 or 0

y x y y y

y y y

v v v v v v v v v v v v

v v v v v v v v v v

Since we are given that B 0v , we must have A 0yv . This means that the final direction of A is

the x direction. Put this result into the momentum equations to find the final speeds.

A A B B A3.7 m s 2.0 m sxv v v v v

Bv

AvA

AvAm

BmBm

Am

B

2

1

Amv

Bmv



180

45. Let A represent the incoming neon atom, and B represent the target atom. A momentum diagram of the collision looks like the first figure. The figure can be re-drawn as a triangle, the second figure, since

A A A A B Bm m mv v v . Write the law of sines for this triangle, relating

each final momentum magnitude to the initial momentum magnitude.

A AA A

A A

B B AB A

A A B

sin sin

sin sin

sin sin

sin sin

m vv v

m v

m v mv v

m v m

The collision is elastic, so write the KE conservation equation, and substitute the results from above. Also note that o o o180.0 55.6 50.0 74.4

22

2 2 2 2 A1 1 1A A A A B B A A A A B A2 2 2

B

sin sin

sin sin

mm v m v m v m v m v m v

m

2 o2

AB 2 2 2 2 o

20.0 u sin 55.6sin39.9 u

sin sin sin 74.4 sin 50.0

mm

46. Use Eq. 7-9a, extended to three particles.

A A B B C CCM

A B C

1.00 kg 0 1.50 kg 0.50 m 1.10 kg 0.75 m

1.00 kg 1.50 kg 1.10 kg

0.44 m

m x m x m xx

m m m

47. Choose the carbon atom as the origin of coordinates. 10

11C C O OCM

C O

12 u 0 16 u 1.13 10 m6.5 10 m

12 u 16 u

m x m xx

m m from the C atom.

48. Find the CM relative to the front of the car.

car car front front back backCM

car front back

1050 kg 2.50 m 2 70.0 kg 2.80 m 3 70.0 kg 3.90 m

2.74 m1050 kg 2 70.0 kg 3 70.0 kg

m x m x m xx

m m m

49. Consider this diagram of the cars on the raft. Notice that the origin of coordinates is located at the CM of the raft. Reference all distances to that location.

1200 kg 9 m 1200 kg 9 m 1200 kg 9 m1.04 m

3 1200 kg 6800 kg

1200 kg 9 m 1200 kg 9 m 1200 kg 9 m1.04 m

3 1200 kg 6800 kg

CM

CM

x

y

Av

o50.0

o55.6Av

AmBm

Am

Bm

Bv

A Am v

B Bm vA Am v

x

y



181

50. By the symmetry of the problem, since the centers of the cubes are along a straight line, the vertical CM coordinate will be 0, and the depth CM coordinate will be 0. The only CM coordinate to calculate is the one along the straight line joining the centers. The mass of each cube will be the

volume times the density, and so 3 3 3

1 0 2 0 3 0 , 2 , 3m l m l m l . Measuring from the

left edge of the smallest block, the locations of the CM s of the individual cubes are 1

1 0 2 0 3 02

, 2 , 4.5x l x l x l . Use Eq. 7-9a to calculate the CM of the system.

3 3 310 0 0 0 0 021 1 2 2 3 3

CM 3 3 3

1 2 3 0 0 0

0

8 2 27 4.5

8 27

3.8 from the left edge of the smallest cube

l l l l l lm x m x m xx

m m m l l l

l

51. Let each crate have a mass M. A top view of the pallet is shown, with the total mass of each stack listed. Take the origin to be the back left corner of the pallet.

5 2 3 3 2 2 5 21.2

10

7 2 2 3 2 1 5 20.9

10

CM

CM

M l M l M lx l

M

M l M l M ly l

M

52. Consider the following. We start with a full circle of radius 2R, with its CM at the origin. Then we draw a circle of radius R, with its CM at the coordinates 0.80 ,0R .

The full circle can now be labeled as a gray part and a white part. The y coordinate of the CM of the entire circle, the CM of the gray part, and the CM of the white part are all at 0y by the symmetry of the system. The x coordinate of the

entire circle is at CM 0x , and can be calculated by gray gray white white

CM

total

m x m xx

m. Rearrange this

equation.

gray gray white white

CM

total

total CM white white total CM white white white whitegray

gray total white total white

m x m x

xm

m x m x m x m x m xx

m m m m m

This is functionally the same as treating the white part of the figure as a hole of negative mass. The mass of each part can be found by multiplying the area of the part times the uniform density of the plate.

2

white whitegray 2 2

total white

0.80 0.800.27

32

R Rm x Rx R

m m R R

53. Take the upper leg, lower leg, and foot all together. Note that Table 7-1 gives the relative mass of BOTH legs and feet, so a factor of 1/2 is needed. Assume a person of mass 70 kg.

21.5 9.6 3.4 170 kg 12 kg

100 2.

2M

2M

3M

M

M

M

x

y



182

54. With the shoulder as the origin of coordinates for measuring the center of mass, we have the following relative locations from Table 7-1 for the arm components, as percentages of the height. Down is positive.

upper lower handarm arm

81.2 7.7 9.5 81.2 55.3 25.9 81.2 43.1 38.1x x x

To find the CM, we can also use relative mass percentages. Since the expression includes the total mass in the denominator, there is no need to divide all masses by 2 to find single component masses. Simply use the relative mass percentages given in the table.

upper upper lower lower hand handarm arm arm arm

CM


9.5 6.6 25.9 4.2 38.1 1.7

6.6 4.2 1.7

19% of the person's height along the line from the shoulder to the hand

x m x m x m

xm m m

55. Take the shoulder to be the origin of coordinates. We assume that the arm is held with the upper arm parallel to the floor and the lower arm and hand extended upward. Measure x horizontally from the shoulder, and y vertically. Since the expression includes the total mass in the denominator, there is no need to divide all masses by 2 to find single component masses. Simply use the relative mass percentages given in the table.


CM


81.2 71.7 6.6 81.2 62.2 4.2 1.7

14.06.6 4.2 1.7

x m x m x m

xm m m


CM


0 6.6 62.2 55.3 4.2 62.2 43.1 1.7

4.926.6 4.2 1.7

y m y m y m

ym m m

Convert the distance percentages to actual distance using the person s height.

CM CM14.0% 155 cm 21.7 cm 4.92% 155 cm 7.6 cmx y

56. See the diagram of the person. The head, trunk, and neck are all lined up so that their CM s are on the torso s median line. Call down the positive y direction. The y distances of the CM of each body part from the median line, in terms of percentage of full height, are shown below, followed by the percentage each body part is of the full body mass.

On median line: head (h): 0 ; 6.9% body mass Trunk & neck (t n): 0 ; 46.1% body mass

From shoulder hinge point: upper arms (u a): 81.2 71.7 = 9.5 ; 6.6% body mass lower arms (l a): 81.2 55.3 = 25.9 ; 4.2% body mass hands (ha): 81.2 43.1 = 38.1 ; 1.7% body mass

trunk and neck

head

feet

lower arms

hands

upper arms

lower legs

upper legs



183

From hip hinge point: upper legs (u l): 52.1 42.5 = 9.6 ; 21.5% body mass lower legs (l l): 52.1 18.2 = 33.9 ; 9.6% body mass feet (f): 52.1 1.8 = 50.3 ; 3.4% body mass

Using this data, calculate the vertical location of the CM.

h h t n t n u a u a l a l a ha ha u l u l l l l l f f

fullbody

0 0 9.5 6.6 25.9 4.2 38.1 1.7 9.6 21.5 33.9 9.6 50.3 3.4

100

9.4

CM

y m y m y m y m y m y m y m y my

m

Thus the center of mass is 9.4% of the full body height below the torso s median line. For a person of height 1.7 m, this is about 16 cm. That is most likely slightly outside the body .

57. (a) Find the CM relative to the center of the Earth. 24 22 8

24 22

6

5.98 10 kg 0 7.35 10 kg 3.84 10 m

5.98 10 kg 7.35 10 kg

4.66 10 m from the center of the Earth

E E M MCM

E M

m x m xx

m m

This is actually inside the volume of the Earth, since 66.38 10 mER

(b) It is this Earth Moon CM location that actually traces out the orbit as discussed in chapter 5. The Earth and Moon will orbit about this location in (approximately) circular orbits. The motion of the Moon, for example, around the Sun would then be a sum of two motions: i) the motion of the Moon about the Earth Moon CM; and ii) the motion of the Earth Moon CM about the Sun. To an external observer, the Moon s motion would appear to be a small radius, higher frequency circular motion (motion about the Earth Moon CM) combined with a large radius, lower frequency circular motion (motion about the Sun).

58. (a) Measure all distances from the original position of the woman.

W W M MCM

W M

55 kg 0 80 kg 10.0 m5.9 m from the woman

135 kg

m x m xx

m m

(b) Since there is no force external to the man-woman system, the CM will not move, relative to the original

position of the woman. The woman s distance will no longer be 0, and the man s distance has changed to 7.5 m.

WW W M MCM

W M

W

M W

55 kg 80 kg 7.5 m5.9 m

135 kg

5.9 m 135 kg 80 kg 7.5 m3.6 m

55 kg

7.5 m 3.6 m 3.9 m

xm x m xx

m m

x

x x

(c) When the man collides with the woman, he will be at the original location of the center of mass.

M Mfinal initial

5.9 m 10.0 m 4.1 mx x

He has moved 4.1 m from his original position.



184

59. The point that will follow a parabolic trajectory is the center of mass. Find the CM relative to the bottom of the mallet. Each part of the hammer (handle and head) can be treated as a point mass located at the CM of the respective piece. So the CM of the handle is 12.0 cm from the bottom of the handle, and the CM of the head is 28.0 cm from the bottom of the handle.

handle handle head headCM

handle head

0.500 kg 24.0 cm 2.00 kg 28.0 cm24.8 cm

2.50 kg

m x m xx

m m

Note that this is inside the head of the mallet.

60. The CM of the system will follow the same path regardless of the way the mass splits, and so will still be 2d from the launch point when the parts land. Assume that the explosion is designed so that

Im still is stopped in midair and falls straight down.

(a) I I II II I I II II 7CM II 3

I II I

3 3 2

4 4

m x m x m d m x d xx d x d

m m m

(b) I I II II II II II IICM II

I II II

3 3

2 54 4

m x m x m d m x d xx d x d

m m m

61. Call the origin of coordinates the CM of the balloon, gondola, and person at rest. Since the CM is at rest, the total momentum of the system relative to the ground is 0. The man climbing the rope cannot change the total momentum of the system, and so the CM must stay at rest. Call the upward direction positive. Then the velocity of the man with respect to the balloon is v . Call the velocity of the balloon with respect to the ground BGv . Then the velocity of the man with respect to the

ground is MG BGv v v . Apply Eq. 7-10.

MG BG BG BG BG0 , upwardm

mv Mv m v v Mv v vm M

If the passenger stops, the balloon also stops, and the CM of the system remains at rest.

62. To find the average force, divide the change in momentum by the time over which the momentum changes. Choose the x direction to be the opposite of the baseball s incoming direction. The velocity with which the ball is moving after hitting the bat can be found from conservation of energy, and knowing the height the ball rises.

21afterinitial final 2collision

2 2 2 9.80 m s 55.6 m 33.0 m s

KE PE mv mg y

v g y

The average force can be calculated from the change in momentum and the time of contact.

3

3

3

3

2 2 3 1 o

0.145 kg 0 35.0 m s3.6 10 N

1.4 10 s

0.145 kg 33.0 m s 03.4 10 N

1.4 10 s

5.0 10 N tan 43

x xxx

y yy

y

y

x y

x

m v vpF

t t

m v vpF

t t

FF F F

F

y

x

vv



185

63. Momentum will be conserved in two dimensions. The fuel was ejected in the y direction as seen from the ground, and so the fuel had no x-component of velocity.

32rocket 0 rocket fuel fuel rocket 03 2

: 0 x x x xp m v m m v m m v v v

1 2

fuel fuel rocket fuel rocket 0 rocket 03 3: 0 2 y y y yp m v m m v m v m v v v

64. In an elastic collision between two objects of equal mass, with the target

object initially stationary, the angle between the final velocities of the objects is 90o. Here is a proof of that fact. Momentum conservation as a vector relationship says A B A B

mv mv mv v v v . Kinetic energy

conservation says 2 2 2 2 2 21 1 1A B A B2 2 2

mv mv mv v v v . The vector

equation resulting from momentum conservation can be illustrated by the second diagram. Apply the law of cosines to that triangle of vectors, and then

equate the two expressions for 2v . 2 2 2

A B A B2 cosv v v v v

Equating the two expressions for 2v gives 2 2 2 2 o

A B A B A B2 cos cos 0 90v v v v v v

For this specific circumstance, see the third diagram. We assume that the target ball is hit correctly so that it goes in the pocket. Find 1 from

the geometry of the left triangle: 1 o

1

1.0tan 30

3.0. Find 2 from

the geometry of the right triangle: 1 o

2

3.0tan 60

3.0. Since the

balls will separate at a 90o angle, if the target ball goes in the pocket, this does appear to be a

good possibility of a scratch shot .

65. (a) The momentum of the astronaut space capsule combination will be conserved since the only forces are internal to that system. Let A represent the astronaut and B represent the space capsule, and let the direction the astronaut moves be the positive direction. Due to the choice of reference frame, A B 0v v . We also have A 2.50 m sv .

initial final A A B A A B B

AB A

B

0

140 kg2.50 m s 0.194 m s

1800 kg

Bp p m v m v m v m v

mv v

m

The negative sign indicates that the space capsule is moving in the opposite direction to the astronaut.

(b) The average force on the astronaut is the astronaut s change in momentum, divided by the time of interaction.

2A A 140 kg 2.50 m s 08.8 10 N

0.40 s

m v vpF

t t

1.0 m

3.0 m

3.0 m

12

v

BvAv

m

vm

m

m

Av

Bv



186

66. Since the only forces on the astronauts are internal to the 2-astronaut system, their CM will not change. Call the CM location the origin of coordinates. That is also the original location of the two astronauts.

BA A B BCM

A B

60 kg 12 m 80 kg

0 9 m

140 kg

xm x m xx x

m m

Their distance apart is A B 12 m 9m 21 mx x .

67. Let A represent the incoming ball, and B represent the target ball. We have B 0v and 1A A4

v v .

Use Eq. 7-7 to obtain a relationship between the velocities. 3

A B A B B A A A4 v v v v v v v v

Substitute this relationship into the momentum conservation equation for the collision. 3 51

initial final A A A A B B A A B A B A4 4 3 p p m v m v m v m v m v m m

68. We assume that all motion is along a single direction. The distance of sliding can be related to the change in the kinetic energy of a car, as follows.

2 2 o1fr fr fr N2

2 212

cos180 f i k k

k f i

W KE m v v W F x F x mg x

g x v v

For post-collision sliding, 0fv and iv is the speed immediately after the collision, v . Use this

relationship to find the speed of each car immediately after the collision.

Car A: 2 21A A A A2

2 2 0.60 9.8m s 18 m 14.55m sk kg x v v g x

Car B: 2 21B B B B2

2 2 0.60 9.8m s 30 m 18.78m sk kg x v v g x

During the collision, momentum is conserved in one dimension. Note that B 0v .

initial final A A A A B B

A A B B

A

1900 kg 14.55m s 1100 kg 18.78m s25.42 m s

1900 kgA

p p m v m v m v

m v m vv

m

For pre-collision sliding, again apply the friction energy relationship, with Afv v and iv is the

speed when the brakes were first applied. 22 2 2 21

A A2

2 25.42 m s 2 0.60 9.8m s 15 m

1mi h28.68m s 64mi h

0.447 m s

k i i A k Ag x v v v v g x

69. Because all of the collisions are perfectly elastic, no energy is lost in the collisions. With each collision, the horizontal velocity is constant, and the vertical velocity reverses direction. So, after each collision, the ball rises again to the same height from which it dropped. Thus, after five bounces, the bounce height will be 4.00 m, the same as the starting height.

70. This is a ballistic pendulum of sorts, similar to Example 7-10 in the textbook. There is no difference in the fact that the block and bullet are moving vertically instead of horizontally. The collision is still totally inelastic and conserves momentum, and the energy is still conserved in the



187

rising of the block and embedded bullet after the collision. So we simply quote the equation from that example.

22

2

2

0.0290 kg 510 m s1 15.47 m

2 0.0290 kg 1.40 kg2 9.80 m s

m Mv gh

m

mvh

g m M

71. This is a ballistic pendulum of sorts, similar to Example 7-10 in the textbook. Momentum is conserved in the totally inelastic collision, and so mv m M v . The kinetic energy present

immediately after the collision is lost due to negative work being done by friction. 2 2 o1

afterfr fr fr N2collision

2 2 21 12 2

cos180

2

f i k k

k f i k

W KE m v v W F x F x mg x

g x v v v v g x

Use this expression for v in the momentum equation in order to solve for v .

2 2

2

0.025 kg 1.35 kg2 2 0.25 9.8 m s 9.5 m 3.8 10 m s

0.025 kg

k

k

mv m M v m M g x

m Mv g x

m

72. Calculate the CM relative to the 60-kg person s seat, at one end of the boat. See the first diagram. Don t forget to include the boat s mass.

A A B B C CCM

A B C

60 kg 0 80 kg 1.6 m 75 kg 3.2 m

1.712 m215 kg

m x m x m xx

m m m

Now, when the passengers exchange positions, the boat will move some distance d as shown, but the CM will not move. We measure the location of the CM from the same place as before, but now the boat has moved relative to that origin.

A A B B C CCM

A B C

75 kg 80 kg 1.6 m 60 kg 3.2 m 215 kg m 320 kg m1.712 m

215 kg 215 kg

0.224 m

m x m x m xx

m m m

d d d d

d

Thus the boat will move 0.22 m towards the initial position of the 75 kg person .

73. (a) The meteor striking and coming to rest in the Earth is a totally inelastic collision. Let A represent the Earth and B represent the meteor. Use the frame of reference in which the Earth is at rest before the collision, and so A 0v . Write momentum conservation for the collision.

B B B m v m m v

75 kg 80 kg 60 kg

d

60 kg 80 kg 75 kg



188

84 13B

B 24 8

A B

1.0 10 kg1.5 10 m s 2.5 10 m s

6.0 10 kg 1.0 10 kg

mv v

m m

(b) The fraction of the meteor s KE transferred to the Earth is the final KE of the Earth divided by

the initial KE of the meteor. 224 1312final 1

2Earth 172

221 8 41initial B B2 2meteor

6.0 10 kg 2.5 10 m s1.7 10

1.0 10 kg 1.5 10 m s

KEm v

KE m v

(c) The Earth s change in KE can be calculated directly. 2 24 131 1

Earth final initial 2 2Earth Earth

0 6.0 10 kg 2.5 10 m s 0.19 JKE KE KE m v

74. Momentum will be conserved in one dimension in the explosion. Let A represent the fragment with the larger KE.

A Ainitial final A A B B B

B

2

2 2 A A A1 1A A B B B2 2

B B

0

12 2

2A B

m vp p m v m v v

m

m v mKE KE m v m v m

m m

.

The fragment with the larger KE energy has half the mass of the other fragment.

75. (a) The force is linear, with a maximum force of 580 N at 0 seconds, and a minimum force of 40 N at 3 milliseconds.

(b) The impulse given the bullet is the area under the F vs. t graph. The area is trapezoidal.

3580 N 40 NImpulse 3.0 10 s

2

0.93N s

(c) The impulse given the bullet is the change in momentum of the bullet. The starting speed of the bullet is 0.

3

0

Impulse 0.93 N sImpulse 4.2 10 kg

220 m sp m v v m

v

76. For the swinging balls, their velocity at the bottom of the swing and the height to which they rise are related by conservation of energy. If the zero of gravitational potential energy is taken to be the lowest point of the swing, then the kinetic energy at the low point is equal to the potential energy at

the highest point of the swing, where the speed is zero. Thus we have 21bottom2

mv mgh for any

swinging ball, and so the relationship between speed and height is 2

bottom 2v gh .

(a) Calculate the speed of the lighter ball at the bottom of its swing. 2 o

A A2 2 9.8 m s 0.30 m 0.30 m cos 60 1.715 m s 1.7 m sv gh

(b) Assume that the collision is elastic, and use the results of problem 30. Take the direction that ball A is moving just before the collision as the positive direction.

0

200

400

600

0 0.5 1 1.5 2 2.5 3

milliseconds

New

ton

s



189

A BA A

A B

AB

A B

0.040 kg 0.060 kg1.715 m s 0.343m s 0.34 m s

0.040 kg 0.060 kg

2 0.040 kg21.715 m s 1.372 m s 1.4 m s

0.040 kg 0.060 kgA

m mv v

m m

mv v

m m

Notice that ball A has rebounded backwards.

(c) After each collision, use the conservation of energy relationship again. 2 22 2

3 2A BA B2 2

0.343m s 1.372 m s6.0 10 m 9.6 10 m

2 22 9.8 m s 2 9.8 m s

v vh h

g g

77. Use conservation of momentum in one dimension, since the particles will separate and travel in opposite directions. Let A represent the alpha particle, and B represent the smaller nucleus. Call the direction of the alpha particle s motion the positive direction. We have B A57m m , A B 0v v ,

and 5

A 3.8 10 m sv .

initial final

53 3

0

3.8 10 m s6.7 10 m s , 6.7 10 m s

57 57

A A B B

A A AB B

B

p p m v m v

m v vv v

m

The negative sign indicates that the nucleus is moving in the opposite direction of the alpha particle.

78. The original horizontal distance can be found from the range formula from Example 3-8. 22 o 2

0 0sin 2 25 m s sin 60 9.8 m s 55.2 mR v g

The height at which the objects collide can be found from Eq. 2-11c for the vertical motion, with 0yv at the top of the path. Take up to be positive.

2o2 2

02 2

0 0 0 2

0 25 m s sin 302 7.97 m

2 2 9.8 m sy y

y y

v vv v a y y y y

a

Let m represent the bullet and M the skeet. When the objects collide, the skeet is moving horizontally at o

0 cos 25 m s cos 30 21.65 m s xv v , and the bullet is moving vertically at

200 m syv . Write momentum conservation in both directions to find the velocities after the

totally inelastic collision. 0.25 kg 21.65 m s

: 20.42 m s0.25 0.015 kg

0.015 kg 200 m s: 11.32 m s

0.25 0.015 kg

xx x x x

y

y y y y

Mvp Mv M m v v

M m

mvp mv M m v v

M m

(a) The speed yv can be used as the starting vertical speed in Eq. 2-11c to find the height that the

skeet-bullet combination rises above the point of collision. 22 2

02 2

0 0 0extra extra 2

0 11.32 m s2 6.5 m

2 2 9.8 m sy y

y y

v vv v a y y y y

a

(b) From Eq. 2-11b applied to the vertical motion after the collision, we can find the time for the skeet-bullet combination to reach the ground.



190

2 2 21 10 2 2

2

0 7.97 m 11.32 m s 9.8 m s

4.9 11.32 7.97 0 2.88 s , 0.565 s

yy y v t at t t

t t t

The positive time root is used to find the horizontal distance traveled by the combination after the collision.

after 20.42 m s 2.88 s 58.7 mxx v t

If the collision would not have happened, the skeet would have gone 1

2 R horizontally.

1 1after 2 2

58.7 m 55.2 m 31.1 m 31 mx x R

79. (a) Use conservation of energy to find the speed of mass m before the collision. The potential energy at the starting point is all transformed into kinetic energy just before the collision.

2 21A A A A2

2 2 9.80 m s 3.60 m 8.40 m smgh mv v gh

Use Eq. 7-7 to obtain a relationship between the velocities, noting that 0Bv .

A B B A B A A

v v v v v v v

Apply momentum conservation for the collision, and substitute the result from Eq. 7-7.

A A B A A A

A A

B A A

2.20 kg 7.00 kg8.4 m s 4.38 m s 4.4 m s

9.20 kg

4.4 m s 8.4 m s 4.0 m s

mv mv Mv mv M v v

m Mv v

m M

v v v

(b) Again use energy conservation to find the height to which mass m rises after the collision. The kinetic energy of m immediately after the collision is all transformed into potential energy. Use the angle of the plane to change the final height into a distance along the incline.

22 A1

A A A2

22

A AA 2

2

4.38m s1.96 m 2.0 m

sin 30 2 sin 30 2 9.8m s sin 30

vmv mgh h

g

h vd

g g

80. Let A represent mass m, and B represent mass M. Use Eq. 7-7 to obtain a relationship between the velocities, noting that 0Bv .

A B B A A B A

v v v v v v v .

After the collision, Av will be negative since m is moving in the negative direction. For there to be a

second collision, then after m moves up the ramp and comes back down, with a positive velocity at the bottom of the incline of Av , the speed of m must be greater than the speed of M so that m can

catch M. Thus A Bv v , or A Bv v . Substitute the result from Eq. 7-7 into the inequality. 1

B A B B A2 v v v v v .

Now write momentum conservation for the original collision, and substitute the result from Eq. 7-7.

A A B B A B B A

2 mmv mv Mv m v v Mv v v

m M

Finally, combine the above result with the inequality from above.



191

1 1A A2 3

2

4 2.33 kg

mv v m m M m M

m M

81. The interaction between the planet and the spacecraft is elastic, because the force of gravity is

conservative. Thus kinetic energy is conserved in the interaction. Consider the problem a 1-dimensional collision, with A representing the spacecraft and B representing Saturn. Because the mass of Saturn is so much bigger than the mass of the spacecraft, Saturn s speed is not changed appreciably during the interaction. Use Eq. 7-7, with A 10.4 km sv and B B 9.6 km sv v .

A B A B A B A

2 2 9.6 km s 10.4 km s 29.6 km sv v v v v v v

Thus there is almost a threefold increase in the spacecraft s speed.

CHAPTER 7: Linear Momentum -...

Documents

Transcript of CHAPTER 7: Linear Momentum -...